let-p-x-1-jx-n-1-jx-n-with-j-e-i2pi-3-1-determine-the-roots-of-p-x-and-factorize-P-x-inside-C-x-2-decompose-the-fraction-F-x-1-p-x-

Question Number 74225 by mathmax by abdo last updated on 20/Nov/19

l e t p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n} w i t h j = e^{\frac{i 2 π}{3}}

1) d e t e r m i n e t h e r o o t s o f p (x) a n d f a c t o r i z e P (x) i n s i d e C [x]

2) d e c o m p o s e t h e f r a c t i o n F (x) = \frac{1}{p (x)}

Answered by mind is power last updated on 20/Nov/19

p (x) = 0 \Rightarrow {(1 + j x)}^{n} - {(1 - j x)}^{n} = 0

You can't use 'macro parameter character #' in math mode

\frac{1 + j x}{1 - j x} = e^{\frac{2 i k π}{n}}, x \neq \frac{1}{j} k < n

\Leftrightarrow x = \frac{e^{\frac{2 i k π}{n}} - 1}{j (e^{\frac{2 i k π}{n}} + 1)} = \frac{1}{j} \frac{e^{i \frac{k π}{n}} - e^{\frac{- i k π}{n}}}{e^{i k \frac{π}{n}} + e^{- \frac{i k π}{n}}} = \frac{i t a n (\frac{k π}{n})}{j}

x = \frac{i}{j} . t a n (\frac{k π}{2 n}) k \in [0, n - 1] w i t h e \frac{k}{n} \neq 2

i f n = 2 s \Rightarrow k \in [1, 2 s - 1] - {s}

i f n = 2 s + 1 \Rightarrow k \in [1, 2 s - 1]

p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n}

p (x) = {(j x)}^{n} - {(- j x)}^{n} + n {(j x)}^{n - 1} - n {(- j x)}^{n - 1} \dots \dots . .

p (x) = (j^{n} + {(- 1)}^{n + 1} j^{j}) x^{n} + n (j^{n - 1} + {(- 1)}^{n} j^{n - 1}) x^{n - 1}

i f n = 2 s p i s p o l y n o m d e g p = 2 s - 1

i f n = 2 s + 1 d e g p = 2 s + 1

n = 2 s

p (x) = n (2 j^{n - 1}) \underset{k = 0, k \neq \frac{n}{2}}{\prod^{n - 1}} (X - \frac{i t a n (\frac{k π}{n})}{j})

i f n = 2 s + 1

p (x) = (2 j^{n}) \underset{k = 0}{\prod^{n - 1}} (X - \frac{i t a n (\frac{k π}{n})}{j})

2) \frac{1}{p (x)}

w e d o i t f o r n = 2 k + 1

\frac{1}{p (x)} = \underset{k = 0}{\sum^{n - 1}} \frac{a_{k}}{(X - \frac{i t a n (\frac{k π}{n})}{j})}

You can't use 'macro parameter character #' in math mode

You can't use 'macro parameter character #' in math mode

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Commented by mathmax by abdo last updated on 21/Nov/19

t h a n k x s i r .